Factor Theorem & Remainder Theorem Applications | 因式定理与余数定理的应用

📚 Factor Theorem & Remainder Theorem Applications | 因式定理与余数定理的应用

The Factor Theorem and Remainder Theorem are two of the most powerful tools in IB Mathematics: Analysis and Approaches. They allow us to factorise polynomials, solve equations, and find unknown coefficients without performing laborious long division. Yet many students memorise these theorems without truly understanding how to apply them flexibly in exam questions. This guide breaks down both theorems, their proofs, and the full range of applications you will encounter in your IB exams.

因式定理与余数定理是 IB 数学：分析与方法（Analysis and Approaches）中最强大的两个工具。它们让我们无需进行繁琐的长除法，就能对多项式进行因式分解、解方程以及求未知系数。然而，许多学生只是死记硬背这两个定理，却没有真正理解如何在考试中灵活运用。本指南将详细讲解这两个定理、它们的证明，以及你在 IB 考试中会遇到的全部应用场景。

1. Polynomial Division and the Need for Theorems | 多项式除法与定理的必要性

When a polynomial f(x) is divided by a linear divisor (x − a), the division algorithm states that f(x) = (x − a) · q(x) + r, where q(x) is the quotient and r is a constant remainder (since the divisor is linear, the remainder is always a constant). Computing this quotient through algebraic long division is possible, but it is time-consuming and prone to arithmetic errors.

当一个多项式 f(x) 除以线性因式 (x − a) 时，除法算法表明 f(x) = (x − a) · q(x) + r，其中 q(x) 是商，r 是常数余数（因为除数是一次多项式，余数恒为常数）。通过代数长除法计算这个商是可行的，但耗时且容易出错。

f(x) = (x − a)q(x) + r

The Remainder Theorem eliminates the need to find q(x) entirely. It tells us that the remainder r is simply f(a). This single insight saves significant time in both pure mathematics and applied contexts such as series and approximation problems.

余数定理让我们完全无需找出 q(x)。它告诉我们余数 r 就等于 f(a)。这一关键洞察在纯数学以及级数与近似问题等应用场景中都能节省大量时间。

2. The Remainder Theorem — Statement | 余数定理——内容陈述

Theorem: If a polynomial f(x) is divided by (x − a), the remainder is f(a). In symbols: the remainder r satisfies r = f(a).

定理：如果多项式 f(x) 除以 (x − a)，余数为 f(a)。用符号表示为：余数 r 满足 r = f(a)。

Proof: From division we have f(x) = (x − a)q(x) + r. Substituting x = a into both sides gives f(a) = (a − a)q(a) + r = 0 · q(a) + r = r. Hence r = f(a). ∎

证明：由除法可知 f(x) = (x − a)q(x) + r。将 x = a 代入两边得 f(a) = (a − a)q(a) + r = 0 · q(a) + r = r。因此 r = f(a)。∎

Worked Example: Find the remainder when f(x) = x³ − 4x² + 2x − 3 is divided by x − 2. Instead of long division, compute f(2) = 8 − 16 + 4 − 3 = −7. The remainder is −7.

示例：求 f(x) = x³ − 4x² + 2x − 3 除以 x − 2 的余数。不需要长除法，计算 f(2) = 8 − 16 + 4 − 3 = −7。余数为 −7。

Notice that the remainder can be negative. A common misconception is that a remainder must be positive; in polynomial division, the remainder may be any real number, including zero, positive, or negative values.

注意余数可以是负数。一个常见的误解是余数必须为正；在多项式除法中，余数可以是任何实数，包括零、正数或负数。

3. The Factor Theorem — Statement | 因式定理——内容陈述

Theorem: (x − a) is a factor of the polynomial f(x) if and only if f(a) = 0.

定理：(x − a) 是多项式 f(x) 的因式，当且仅当 f(a) = 0。

This is a direct corollary of the Remainder Theorem: if the remainder is zero, then (x − a) divides f(x) exactly, making it a factor. The “if and only if” nature means the theorem works in both directions — you can use it to verify a factor, or to find roots of an equation.

这是余数定理的直接推论：如果余数为零，则 (x − a) 整除 f(x)，因此它是因式。’当且仅当’的性质意味着该定理双向适用——你可以用它来验证因式，也可以用它来求方程的根。

Worked Example: Show that (x − 3) is a factor of f(x) = x³ − 5x² + 8x − 6. Evaluate f(3) = 27 − 45 + 24 − 6 = 0. Since f(3) = 0, the Factor Theorem confirms that (x − 3) is a factor.

示例：证明 (x − 3) 是 f(x) = x³ − 5x² + 8x − 6 的因式。计算 f(3) = 27 − 45 + 24 − 6 = 0。因为 f(3) = 0，因式定理确认 (x − 3) 是它的因式。

4. Factorising Polynomials Using the Factor Theorem | 利用因式定理分解多项式

To fully factorise a polynomial of degree n, you need to find n linear factors (some of which may repeat). The Factor Theorem tells you each zero corresponds to a factor. A systematic strategy for cubic polynomials f(x) = ax³ + bx² + cx + d is as follows:

要完全分解一个 n 次多项式，你需要找到 n 个线性因式（其中一些可能重复）。因式定理告诉我们每个零点对应一个因式。对于三次多项式 f(x) = ax³ + bx² + cx + d，系统的策略如下：

Use the Rational Root Theorem to list candidate roots: factors of d divided by factors of a.
用有理根定理列出候选根：d 的因子除以 a 的因子。
Test each candidate c by evaluating f(c) until you find one where f(c) = 0.
逐一测试候选值 c，计算 f(c)，直到找到一个使 f(c) = 0 的根。
Perform polynomial division (or synthetic division) to obtain the quotient quadratic.
通过多项式除法（或综合除法）求出商二次式。
Factorise the quadratic using standard methods or the quadratic formula.
用标准方法或二次公式分解该二次式。

Full Example: Factorise f(x) = x³ − 4x² + x + 6. Candidate roots from the Rational Root Theorem: ±1, ±2, ±3, ±6. Test x = 1: f(1) = 1 − 4 + 1 + 6 = 4 ≠ 0. Test x = −1: f(−1) = −1 − 4 − 1 + 6 = 0. So (x + 1) is a factor. Dividing gives x³ − 4x² + x + 6 = (x + 1)(x² − 5x + 6) = (x + 1)(x − 2)(x − 3).

完整示例：分解 f(x) = x³ − 4x² + x + 6。由有理根定理，候选根为 ±1、±2、±3、±6。测试 x = 1：f(1) = 1 − 4 + 1 + 6 = 4 ≠ 0。测试 x = −1：f(−1) = −1 − 4 − 1 + 6 = 0。所以 (x + 1) 是一个因式。作除法得 x³ − 4x² + x + 6 = (x + 1)(x² − 5x + 6) = (x + 1)(x − 2)(x − 3)。

x³ − 4x² + x + 6 = (x + 1)(x − 2)(x − 3)

5. Solving Polynomial Equations | 解多项式方程

Setting a factorised polynomial equal to zero allows immediate solution. If f(x) = (x − r₁)(x − r₂)(x − r₃), then the equation f(x) = 0 has solutions x = r₁, r₂, r₃. This connection between factors and roots is central to sketching polynomial graphs and solving higher-degree equations.

将分解后的多项式置零即可直接求解。如果 f(x) = (x − r₁)(x − r₂)(x − r₃)，则方程 f(x) = 0 的解为 x = r₁、r₂、r₃。因式与根之间的联系是绘制多项式图形和求解高次方程的核心。

Worked Example: Solve x³ − 2x² − 5x + 6 = 0. Test candidates: f(1) = 1 − 2 − 5 + 6 = 0, so x = 1 is a root. Dividing by (x − 1) yields x² − x − 6 = (x − 3)(x + 2). Hence the full factorisation is (x − 1)(x − 3)(x + 2), and the solutions are x = 1, 3, −2.

示例：解方程 x³ − 2x² − 5x + 6 = 0。测试候选根：f(1) = 1 − 2 − 5 + 6 = 0，所以 x = 1 是一个根。除以 (x − 1) 得 x² − x − 6 = (x − 3)(x + 2)。因此完全分解为 (x − 1)(x − 3)(x + 2)，解为 x = 1、3、−2。

For quartic equations, the same principle applies: find one root, reduce to a cubic, find another root, reduce to a quadratic, and solve. This iterative process is efficient and systematic.

对于四次方程，原理相同：找到一个根，降为三次方程，再找一个根，降为二次方程，然后求解。这种迭代过程高效且系统化。

6. Finding Unknown Coefficients | 求未知系数

A classic IB question provides information about factors or remainders and asks you to determine unknown coefficients in the polynomial. Both theorems are essential here — the Remainder Theorem gives equations, and the Factor Theorem allows you to reduce the number of unknown coefficients.

IB 考试中的经典题型是：给出有关因式或余数的信息，要求你确定多项式中的未知系数。两个定理在这里都至关重要——余数定理给出方程，因式定理允许你减少未知系数的个数。

Example 1 (Two unknowns): The polynomial f(x) = x³ + ax² + bx − 6 has a factor (x − 2) and leaves a remainder of 12 when divided by (x + 1). Find a and b.

示例 1（两个未知数）：多项式 f(x) = x³ + ax² + bx − 6 有因式 (x − 2)，且除以 (x + 1) 时余数为 12。求 a 和 b。

Using the Factor Theorem: f(2) = 8 + 4a + 2b − 6 = 0 → 4a + 2b = −2 → 2a + b = −1. Using the Remainder Theorem: f(−1) = −1 + a − b − 6 = 12 → a − b = 19. Solving simultaneously: adding the two equations gives 3a = 18, so a = 6 and b = −13.

由因式定理：f(2) = 8 + 4a + 2b − 6 = 0 → 4a + 2b = −2 → 2a + b = −1。由余数定理：f(−1) = −1 + a − b − 6 = 12 → a − b = 19。联立求解：两式相加得 3a = 18，所以 a = 6，b = −13。

Example 2 (One unknown): Given that (x + 3) is a factor of g(x) = 2x³ + 7x² + px − 6, find p. Since g(−3) = 2(−27) + 7(9) + p(−3) − 6 = −54 + 63 − 3p − 6 = 3 − 3p = 0, we get p = 1. Always define and solve the equation carefully to avoid sign errors.

示例 2（一个未知数）：已知 (x + 3) 是 g(x) = 2x³ + 7x² + px − 6 的因式，求 p。因为 g(−3) = 2(−27) + 7(9) + p(−3) − 6 = −54 + 63 − 3p − 6 = 3 − 3p = 0，所以 p = 1。务必仔细建立方程并求解，避免符号错误。

7. Division by Expressions of the Form ax + b | 除以 ax + b 形式的表达式

When the divisor is not monic — that is, of the form ax + b rather than x − a — the Remainder Theorem requires adaptation. If f(x) is divided by (ax + b), the remainder is f(−b/a). You must solve ax + b = 0 to find the value to substitute.

当除数不是首一多项式时——即除数为 ax + b 而非 x − a 时——余数定理需要调整。如果 f(x) 除以 (ax + b)，余数为 f(−b/a)。你必须先解方程 ax + b = 0 找到要代入的值。

Worked Example: Find the remainder when f(x) = 3x³ − 2x² + 5x − 1 is divided by 2x − 1. Set 2x − 1 = 0, giving x = ½. Compute f(½) = 3(⅛) − 2(¼) + 5(½) − 1 = ⅜ − ½ + ⁵⁄₂ − 1 = ⅜ − ⁴⁄₈ + ²⁰⁄₈ − ⁸⁄₈ = ¹¹⁄₈.

示例：求 f(x) = 3x³ − 2x² + 5x − 1 除以 2x − 1 的余数。令 2x − 1 = 0，得 x = ½。计算 f(½) = 3(⅛) − 2(¼) + 5(½) − 1 = ⅜ − ½ + ⁵⁄₂ − 1 = ⅜ − ⁴⁄₈ + ²⁰⁄₈ − ⁸⁄₈ = ¹¹⁄₈。

Remainder = f(−b/a) = f(−1/2) for divisor 2x + 1

Be careful with the sign: for divisor ax + b, the value is x = −b/a, not x = b/a. A negative sign error here is the most common mistake in this type of question.

注意符号：对于除数 ax + b，代入值为 x = −b/a，而不是 x = b/a。负号错误是这类题目中最常见的失误。

8. Repeated Factors and Multiplicity | 重因式与重数

Sometimes a factor appears more than once. If (x − a)² is a factor of f(x), then f(a) = 0 and the derivative f'(a) = 0 as well. In IB courses, you may be asked to find unknown coefficients when a polynomial has a repeated root, which requires using both f(a) = 0 and f'(a) = 0.

有时一个因式会出现多次。如果 (x − a)² 是 f(x) 的因式，则 f(a) = 0 且导数 f'(a) = 0。在 IB 课程中，你可能会被要求在多边形具有重根时求未知系数，这需要同时使用 f(a) = 0 和 f'(a) = 0。

Worked Example: Given that (x − 1)² is a factor of h(x) = x³ + ax² + bx − 3, find a and b. We have h(1) = 1 + a + b − 3 = 0 → a + b = 2. Also h'(x) = 3x² + 2ax + b, so h'(1) = 3 + 2a + b = 0 → 2a + b = −3. Solving: subtracting gives a = −5 and b = 7.

示例：已知 (x − 1)² 是 h(x) = x³ + ax² + bx − 3 的因式，求 a 和 b。我们有 h(1) = 1 + a + b − 3 = 0 → a + b = 2。又 h'(x) = 3x² + 2ax + b，所以 h'(1) = 3 + 2a + b = 0 → 2a + b = −3。联立相减得 a = −5，b = 7。

The concept of multiplicity also helps when sketching graphs: a factor with even multiplicity touches the x-axis without crossing, while odd multiplicity crosses the axis. Use the Factor Theorem to identify these behaviours quickly.

重数的概念也有助于绘制图形：偶重因式在 x 轴处相切而不穿过，奇重因式则穿过 x 轴。利用因式定理可以快速判断这些行为。

9. Common Exam Questions and Pitfalls | 常见考题与易错点

IB exams frequently combine the Factor and Remainder Theorems with other topics. Below are the typical question formats and the mistakes students make:

IB 考试经常将因式定理和余数定理与其他主题结合。以下是典型的题目形式以及学生常犯的错误：

Question Type	Common Mistake	Correct Approach
Find remainder of f(x) ÷ (x − a)	Using long division unnecessarily	Substitute x = a into f(x)
Find remainder of f(x) ÷ (ax + b)	Substituting x = b/a instead of −b/a	Solve ax + b = 0 first
Show (x − a) is a factor	Dividing fully before testing	Evaluate f(a); if zero, factor confirmed
Find unknown coefficients	Sign errors when substituting negative roots	Write out f(a) = 0 explicitly with brackets

题目类型：求 f(x) ÷ (x − a) 的余数——常见误区：不必要地使用长除法——正确做法：将 x = a 代入 f(x)。求 f(x) ÷ (ax + b) 的余数——常见误区：代入 x = b/a 而非 −b/a——正确做法：先解 ax + b = 0。证明 (x − a) 是因式——常见误区：先做完全除法再测试——正确做法：计算 f(a)，若为零则因式成立。求未知系数——常见误区：代入负根时符号错误——正确做法：用括号明确写出方程 f(a) = 0。

Another frequent trap: when asked to “fully factorise” a polynomial, students often stop after finding one factor. Always continue until all factors are linear (or irreducible quadratics) and check the degree of the product equals the degree of the original polynomial.

另一个常见陷阱：当被要求’完全分解’多项式时，学生常常找到一个因式就停止了。务必继续分解，直到所有因式为一次式（或不可约二次式），并检查因式的乘积次数等于原多项式的次数。

10. Practice Problem Set with Solutions | 练习题目与解答

Work through these problems to consolidate your understanding. Solutions are provided below each question — attempt each one before revealing the answer.

请完成以下题目来巩固理解。每题下方附有解答——请先尝试再查看答案。

Problem 1: The polynomial f(x) = 2x³ − x² + ax + b leaves remainders 3 and 7 when divided by (x − 1) and (x − 2) respectively. Find a and b.

题目 1：多项式 f(x) = 2x³ − x² + ax + b 分别除以 (x − 1) 和 (x − 2) 时余数为 3 和 7。求 a 和 b。

Solution: f(1) = 2 − 1 + a + b = 3 → a + b = 2. f(2) = 16 − 4 + 2a + b = 7 → 2a + b = −5. Subtracting gives a = −7 and b = 9.

解答：f(1) = 2 − 1 + a + b = 3 → a + b = 2。f(2) = 16 − 4 + 2a + b = 7 → 2a + b = −5。相减得 a = −7，b = 9。

Problem 2: Show that (2x − 1) is a factor of f(x) = 2x³ − x² + 4x − 2, and hence factorise f(x) completely.

题目 2：证明 (2x − 1) 是 f(x) = 2x³ − x² + 4x − 2 的因式，并由此完全分解 f(x)。

Solution: f(½) = 2(⅛) − ¼ + 2 − 2 = ¼ − ¼ + 0 = 0, so (2x − 1) is a factor. Synthetic division gives (2x − 1)(x² + 2), and the quadratic x² + 2 is irreducible over the reals. Thus f(x) = (2x − 1)(x² + 2).

解答：f(½) = 2(⅛) − ¼ + 2 − 2 = ¼ − ¼ + 0 = 0，所以 (2x − 1) 是因式。综合除法得 (2x − 1)(x² + 2)，二次式 x² + 2 在实数范围内不可约。因此 f(x) = (2x − 1)(x² + 2)。

Problem 3: Given that (x − 2) is a factor of g(x) = x³ − 7x + 6, find all real roots of g(x) = 0.

题目 3：已知 (x − 2) 是 g(x) = x³ − 7x + 6 的因式，求 g(x) = 0 的所有实根。

Solution: g(2) = 8 − 14 + 6 = 0, confirming the factor. Dividing by (x − 2): x³ − 7x + 6 = (x − 2)(x² + 2x − 3) = (x − 2)(x + 3)(x − 1). Roots: x = 2, −3, 1.

解答：g(2) = 8 − 14 + 6 = 0，确认因式成立。除以 (x − 2)：x³ − 7x + 6 = (x − 2)(x² + 2x − 3) = (x − 2)(x + 3)(x − 1)。根为 x = 2、−3、1。

11. Connection to Sum and Product of Roots | 与根的和积公式的联系

For a cubic polynomial ax³ + bx² + cx + d with roots α, β, γ, Vieta’s formulas give α + β + γ = −b/a, αβ + βγ + γα = c/a, and αβγ = −d/a. The Factor Theorem provides a direct route to these roots, making Vieta’s formulas useful as a check rather than a primary solving tool.

对于根为 α、β、γ 的三次多项式 ax³ + bx² + cx + d，韦达定理给出 α + β + γ = −b/a、αβ + βγ + γα = c/a 和 αβγ = −d/a。因式定理为我们提供了找到这些根的直接途径，使韦达定理更多地用作校验工具而非主要求解手段。

Worked Example: The roots of x³ + px² + qx − 8 = 0 are 1, 2, and r. Using the Factor Theorem: f(1) = 1 + p + q − 8 = 0. Instead of solving immediately, use the product of roots: 1 × 2 × r = 8, so r = 4. Then use sum: 1 + 2 + 4 = 7 = −p, so p = −7. Finally, pairing products: 1·2 + 1·4 + 2·4 = 14 = q. The polynomial is x³ − 7x² + 14x − 8.

示例：方程 x³ + px² + qx − 8 = 0 的根为 1、2、r。利用因式定理：f(1) = 1 + p + q − 8 = 0。不必立即求解，先使用根的乘积：1 × 2 × r = 8，所以 r = 4。再使用和：1 + 2 + 4 = 7 = −p，所以 p = −7。最后，两两乘积之和：1·2 + 1·4 + 2·4 = 14 = q。该多项式为 x³ − 7x² + 14x − 8。

12. Final Tips for IB Exams | IB 考试最终建议

Master the Factor and Remainder Theorems by practising with polynomials of degree 3 and 4, including those with missing terms (e.g., x³ – 4x + 1 has no x² term). Always write out the substitution step clearly — this earns method marks even if the final answer is incorrect. Use a systematic list of candidate roots rather than guessing randomly, and verify each factorisation by expanding or by evaluating at the roots.

掌握因式定理和余数定理的关键是练习三次和四次多项式，包括缺项的多项式（例如 x³ – 4x + 1 没有 x² 项）。始终清晰写出代入步骤——即使最终答案有误，也能获得方法分。系统列出候选根而不是随机猜测，并通过展开或因式处求值来验证每个因式分解的正确性。

Finally, remember that these theorems are not just isolated tools — they underpin graph sketching, solving inequalities, and even calculus problems involving stationary points of polynomial functions. A strong grasp of the Factor and Remainder Theorems will serve you across the entire IB Mathematics: Analysis and Approaches syllabus.

最后，请记住这些定理不仅仅是孤立的工具——它们支撑着图形绘制、不等式求解，甚至涉及多项式函数驻点的微积分问题。扎实掌握因式定理和余数定理将在整个 IB 数学：分析与方法课程中使你受益。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Factor Theorem & Remainder Theorem Applications | 因式定理与余数定理的应用